1985 AHSME Problems/Problem 22

Problem

In a circle with center $O$ , $AD$ is a diameter, $ABC$ is a chord, $BO = 5$ and $\angle ABO = \ \stackrel{\frown}{CD} \ = 60^{\circ}$ . Then the length of $BC$ is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(305)); label("$5$", B--O, dir(O--B)*dir(90)); label("$60^\circ$", dir(185), dir(185)); label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]$

$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \ } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of the above}$

Solution

Since $\angle CAD$ is an angle inscribed in a $60{^\circ}$ arc, we obtain $\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}$ , so $\triangle ABO$ is a $30^{\circ}$ - $60^{\circ}$ - $90^{\circ}$ right triangle. This gives $AO = BO\sqrt{3} = 5\sqrt{3}$ and $AB = 2BO = 10$ , and now since $AD$ is a diameter, $AD = 2AO = 10\sqrt{3}$ . The fact that $AD$ is a diameter also means that $\angle ACD = 90^{\circ}$ , so $\triangle ACD$ is again a $30^{\circ}$ - $60^{\circ}$ - $90^{\circ}$ right triangle, yielding $CD = \frac{1}{2}AD = 5\sqrt{3}$ and $AC = \frac{\sqrt{3}}{2}AD = 15$ , which finally gives $BC = AC-AB = 15-10 = \boxed{\text{(D)} \ 5}$ .

1985 AHSME (Problems • Answer Key • Resources)
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1985 AHSME Problems/Problem 22

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